Birational invariants, purity and the Gersten conjecture Lectures at ...

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of the smooth projective completion of C. The divisibility of CH 0 (C) follows from that
of the group J(k). Thus for our surface S we have CH 0 (S)/n = 0, from which we deduce
CH d−2 (A d−2 × S)/n = 0, hence CH d−2 (V )/n = 0 for all n > 0.
Since the open set U ⊂ X is smooth, as recalled above we have an embedding
H 3 nr(k(X)/k, Z/n) ↩→ H 0 (U, H 3 (Z/n)).
Thus to show that Hnr(k(X)/k, 3 Z/n) is finite, it suffices to show that H 0 (U, H 3 (Z/n))
is finite. Since, as recalled at the beginning of this section, the group H 3 (U, µ ⊗2
n ) is finite
(because k is algebraically closed), by (4.2), finiteness of H 0 (U, H 3 (Z/n)) is equivalent
to finiteness of CH 2 (U)/nCH 2 (U) (note that Z/n ≃ µ ⊗2
n since k is algebraically closed).
The finite and flat morphism p gives rise to morphisms p ∗ : CH d−2 (U) → CH d−2 (V )
and p ∗ : CH d−2 (V ) → CH d−2 (U). The composite map
p ∗ ◦ p ∗ : CH d−2 (U) → CH d−2 (V ) → CH d−2 (U)
is multiplication by N.
The group m CH 2 (U) is a subquotient of the group H 3 (U, µ ⊗2
m ), as recalled above (the
Merkur’ev/Suslin result is used here), and this last group is finite since k is algebraically
closed.
Letting A = CH d−2 (U) and B = CH d−2 (V ), and bearing in mind the vanishing,
hence finiteness of CH d−2 (V )/n = 0 for all n > 0, the finiteness of CH 2 (U)/nCH 2 (U)
now follows from the purely formal lemma :
LEMMA 4.3.2. — Let f : A → B and g : B → A be homomorphisms of abelian groups.
Assume that g ◦ f is multiplication by the positive integer N > 0. Assume that N A and
B/NB are finite. Let n ≥ 0. Then, if B/nB is finite, so is A/nA.
Proof : Let b i ∈ B, i ∈ I be representatives of the finite set I = B/NB. For each i ∈ I
such that g(b i ) belongs to NA, fix an element a i ∈ A with g(b i ) = Na i . For any a ∈ A we
may find b ∈ B and i ∈ I such that f(a) = Nb + b i . Now Na = g ◦ f(a) = Ng(b) + g(b i ).
Hence a − g(b) − a i belongs to the the N-torsion subgroup N A, which is finite. We
conclude that the quotient C = A/g(B) is a finitely generated abelian group. For any
positive integer n, we have an exact sequence
hence A/nA is finite if B/nB is finite.
B/nB −→ A/nA −→ C/nC −→ 0
It remains to prove the finiteness of CH 2 (X)/n. The argument given above shows
that there exists a nonempty open set U ⊂ X such that CH 2 (U)/n is finite. Let F ⊂ X
be the complement of U in X. We have the localization sequence
CH d−2 (F ) → CH d−2 (X) → CH d−2 (U) → 0.
Since dim(F ) ≤ d − 1, the finiteness of CH d−2 (X)/n follows from the lemma :